$g(n) = -3n^{2}+5(h(n))$ $f(x) = 4x^{3}+4x+6-3(g(x))$ $h(t) = -5t^{2}+7t$ $ h(f(1)) = {?} $
Answer: First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = 4(1^{3})+(4)(1)+6-3(g(1))$ To solve for the value of $f$ , we need to solve for the value of $g(1)$ $g(1) = -3(1^{2})+5(h(1))$ To solve for the value of $g$ , we need to solve for the value of $h(1)$ $h(1) = -5(1^{2})+(7)(1)$ $h(1) = 2$ That means $g(1) = -3(1^{2})+(5)(2)$ $g(1) = 7$ That means $f(1) = 4(1^{3})+(4)(1)+6+(-3)(7)$ $f(1) = -7$ Now we know that $f(1) = -7$ . Let's solve for $h(f(1))$ , which is $h(-7)$ $h(-7) = -5(-7)^{2}+(7)(-7)$ $h(-7) = -294$